Question:

Step 1: Determine the total number of possible arrangements. There are 6 students and 6 seats. The total number of ways to allot seats randomly is the number of permutations of 6 students in 6 seats, which is $6!$. $$ \text{Total arrangements} = 6! = 720 $$ Step 2: Evaluate the probability for statement (a). Statement (a) asks for the probability that student $S_4$ gets seat $R_4$, and none of the remaining 5 students gets the seat previously allotted to them. If $S_4$ gets $R_4$, there is only 1 way for this specific assignment. The remaining 5 students ($S_1, S_2, S_3, S_5, S_6$) must be seated in the remaining 5 seats ($R_1, R_2, R_3, R_5, R_6$) such that none of them occupies their original seat. This is a derangement of 5 elements, denoted as $D_5$. The number of derangements of $n$ elements is given by $D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}$. For $n=5$: $$ D_5 = 5! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right) $$ $$ D_5 = 120 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right) $$ $$ D_5 = 60 - 20 + 5 - 1 = 44 $$ The number of favorable outcomes is $1 \times D_5 = 44$. The probability is the ratio of favorable outcomes to total arrangements: $$ P(\text{a}) = \frac{44}{720} = \frac{11}{180} $$ Step 3: Evaluate the probability for statement (b). Statement (b) asks for the probability that student $S_5$ gets seat $R_5$, and none of the remaining 5 students gets the seat previously allotted to them. This scenario is identical in structure to statement (a). The number of favorable outcomes is $D_5 = 44$. The probability is: $$ P(\text{b}) = \frac{44}{720} = \frac{11}{180} $$ Step 4: Evaluate the probability for statements (c) and (d). Statements (c) and (d) ask for the probability that even-numbered students are seated at even-numbered seats, and none of the students is seated on the seat previously allotted. Let $S_E = \{S_2, S_4, S_6\}$ be the set of even-numbered students and $R_E = \{R_2, R_4, R_6\}$ be the set of even-numbered seats. Let $S_O = \{S_1, S_3, S_5\}$ be the set of odd-numbered students and $R_O = \{R_1, R_3, R_5\}$ be the set of odd-numbered seats. The conditions are: 1. Even-numbered students are seated in even-numbered seats. This implies that odd-numbered students must be seated in odd-numbered seats. * The number of ways to arrange $S_E$ in $R_E$ is $3!$. * The number of ways to arrange $S_O$ in $R_O$ is $3!$. The total number of arrangements satisfying this partitioning is $3! \times 3! = 6 \times 6 = 36$. 2. None of the students is seated on their previously allotted seat. This means $S_i \neq R_i$ for all $i \in \{1, 2, 3, 4, 5, 6\}$. * For the even-numbered students ($S_2, S_4, S_6$) in even-numbered seats ($R_2, R_4, R_6$), none can be in their original seat. This is a derangement of 3 elements, $D_3$. $$ D_3 = 3! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} \right) = 6 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} \right) = 3 - 1 = 2 $$ * Similarly, for the odd-numbered students ($S_1, S_3, S_5$) in odd-numbered seats ($R_1, R_3, R_5$), none can be in their original seat. This is also a derangement of 3 elements, $D_3 = 2$. The number of favorable outcomes satisfying both conditions is $D_3 \times D_3 = 2 \times 2 = 4$. The probability is the ratio of favorable outcomes to total arrangements: $$ P(\text{c, d}) = \frac{4}{720} = \frac{1}{180} $$